Question: Solve for $x$, ignoring any extraneous solutions: $\dfrac{x^2 - 7}{x + 5} = \dfrac{-3x + 3}{x + 5}$
Explanation: Multiply both sides by $x + 5$ $ \dfrac{x^2 - 7}{x + 5} (x + 5) = \dfrac{-3x + 3}{x + 5} (x + 5)$ $ x^2 - 7 = -3x + 3$ Subtract $-3x + 3$ from both sides: $ x^2 - 7 - (-3x + 3) = -3x + 3 - (-3x + 3)$ $ x^2 - 7 + 3x - 3 = 0$ $ x^2 - 10 + 3x = 0$ Factor the expression: $ (x - 2)(x + 5) = 0$ Therefore $x = 2$ or $x = -5$ However, the original expression is undefined when $x = -5$. Therefore, the only solution is $x = 2$.